2x^2+26=20x

Solution for 2x^2+26=20x equation:

2x^2+26=20x

We move all terms to the left:

2x^2+26-(20x)=0

a = 2; b = -20; c = +26;
Δ = b2-4ac
Δ = -202-4·2·26
Δ = 192
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-8\sqrt{3}}{2*2}=\frac{20-8\sqrt{3}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+8\sqrt{3}}{2*2}=\frac{20+8\sqrt{3}}{4}
$